# a story about special relativity,who can explain it？

3rd case is beyond me. It isn't a simple formula. There are whole papers written on the subject.

If the weakest gravitational wave could rise the fluctuation of the LIGO interference fringe of light, how should we neglect the gravity’s influences on light?

If light is held by gravity, then measuring the speed of light anywhere on the earth will be a constant.

If classical physics can explain the Morley experiment, it can explain the bending of light, and it can deduce the mass energy equation. So is it necessary to assume that the speed of light is constant ?

So is it necessary to assume that the speed of light is constant ?
I believe it depends on the medium.
Does the speed of light change in air or water?
Yes. Light is slowed down in transparent media such as air, water and glass. The ratio by which it is slowed is called the refractive index of the medium and is usually greater than one.* This was discovered by Jean Foucault in 1850.

When people talk about "the speed of light" in a general context, they usually mean the speed of light in a vacuum. They also usually mean the speed as measured in an inertial frame. This vacuum-inertial speed is denoted c.
http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/speed_of_light.html

But this problem is so simple for classical physics.
The web sites make it look really complicated, but my intuitions generate the following:
Rotate the picture 45 degrees. If Earth is moving at .9c diagonally, it must have component speed of .6364c, right?
Similar, B has the same component speed in the final direction of motion.
Add two of those using the above formula and you get .9059c.

What's so complicated about that? Maybe I made some naive mistake. The web sites make it sound like more effort is needed. Maybe its only complicated if the two speeds being added are not identical, but even then it seems the technique I just used would be valid.

If the weakest gravitational wave could rise the fluctuation of the LIGO interference fringe of light, how should we neglect the gravity’s influences on light?
Well, you can't ignore gravity, but there is no assumed significant gravity in your scenario.

If light is held by gravity, then measuring the speed of light anywhere on the earth will be a constant.
Anywhere period, if a local measurement. Earth to moon is not local, so light speed is a bit above c if measured as a trip to the moon and back, but if the same experiment is done on the moon with a reflector on Earth, you get a value less than c. All this is because of the gravity well depth difference.

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I have made a detailed analysis in several parts:
• Impact of water on vehicle speed
• Impact of air on light propagation
• The influence of the gravitational field on the propagation of light
• Eddington observed the solar eclipse to verify general relativity
• Mass energy equation
• Fiber optic gyroscope
• Explanation of Sagnac effect
• Ask questions to special relativity

Ah, a denier. Your paper has many classic mistakes.

The Sagnac section for instance has formulas that posit different light speed for the different directions, instead of noting that the light has more distance to travel with the spin than it does against it. It is this difference of distance that is measured by fiberoptic gyroscopes that utilize the effect for their measurements.

The tail end of the page ends with a scenario that "cannot be explained by SR". Not sure why you'd assert that. Just because you're not up to it doesn't mean the theory fails there. It's actually a completely trivial scenario:

1) Yes, B1 still reaches X in 4.36 seconds according to its own clock. This is true in any frame.
2) All three ships are moving at .9c relative to Earth, so during 10 Earth seconds, in Earth frame, 4.36 seconds pass in any ship moving with that relative speed.
I say 'in Earth frame' since the frame of the question is left unspecified, and the velocities are given in no one consistent frame, and hence the question is technically ambiguously worded.

They all indeed have easy answers from classic Newtonian theory, which is 10 seconds for everybody. This is demonstrably incorrect, but Newton's theory also allows ships to move faster than light, just like airplanes can move faster than sound.

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Ah, a denier. Your paper has many classic mistakes.

The Sagnac effect for instance has formulas that posit different light speed for the different directions, instead of noting that the light has more distance to travel with the spin than it does against it. It is this difference of distance that is measured by fiberoptic gyroscopes that utilize the effect for their measurements.

The tail end of the page ends with a scenario that "cannot be explained by SR". Not sure why you'd assert that. Just because you're not up to it doesn't mean the theory fails there. It's actually a completely trivial scenario:

1) Yes, B1 still reaches X in 4.36 seconds according to its own clock. SR demands this.
2) All three ships are moving at .9c relative to Earth, so during 10 Earth seconds, in Earth frame, 4.36 seconds pass in any ship moving with that relative speed.

They all indeed have easy answers from classic Newtonian theory, which is 10 seconds for everybody. This is demonstrably incorrect, but Newton's theory allows ships to move faster than light, just like airplanes can move faster than sound.
You can challenge my third question.

Wrong assumptions often lead to right results. For example: 1 - 9 = 7, then 1 - 9 + 9 = ?
I believe that someone can prove that the result is 1, 1 + (9 - 9) = 1.

It is easy to derive the mass energy equation in classical physics.
The special theory of relativity also deduces the equation of mass and energy, which does not mean that this assumption is correct.

A nit on the notation used: You write .9ls (lowercase .9 light seconds), but given the font used, it looks exactly like 0.91 s (about 10/11th of a second, a time interval instead of a distance). Use LS or Ls for light seconds, or at least put a space after the 9.
Also, I think you didn't mean the put the decimal in that speed in the 2nd to last picture. It's 0.9 in one pic, and 9 in the other.

Just a little constructive criticism to make a paper slightly less confusing, albeit still wrong.

A nit on the notation used: You write .9ls (lowercase .9 light seconds), but given the font used, it looks exactly like 0.91 s (about 10/11th of a second, a time interval instead of a distance). Use LS or Ls for light seconds, or at least put a space after the 9.
Also, I think you didn't mean the put the decimal in that speed in the 2nd to last picture. It's 0.9 in one pic, and 9 in the other.

Just a little constructive criticism to make a paper slightly less confusing, albeit still wrong.
Thank you very much. It's LS，I will revise the article. 9LS is right.

Revised article：

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A nit on the notation used: You write .9ls (lowercase .9 light seconds), but given the font used, it looks exactly like 0.91 s (about 10/11th of a second, a time interval instead of a distance). Use LS or Ls for light seconds, or at least put a space after the 9.
Also, I think you didn't mean the put the decimal in that speed in the 2nd to last picture. It's 0.9 in one pic, and 9 in the other.
Just a little constructive criticism to make a paper slightly less confusing, albeit still wrong.
I have changed the editorial errors you pointed out.
So let's talk about our key points. Where do you think it is, which statement is wrong, which makes you confused. You can refute it, and I will be happy to answer your question.
Generally speaking, an article is wrong and irresponsible. Now that I have published the article, I can accept your questions and give you my answers.

At the same time, I'm looking forward to your answering the third question.

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I have changed the editorial errors you pointed out.
Now that I have published the article, I can accept your questions and give you my answers.

At the same time, I'm looking forward to your answering the third question.
I pointed out several things in post 25. The beginning starts off with some kind of analogy of a car on a road with water on it. Light, to the surprise of the scientists of the late 19th century, does not behave like it travels through a medium, so not sure what all that was trying to prove. I pretty much skipped to the bottom where I found bad physics concerning fiberoptic gyroscopes and then a section at the bottom that asserted that SR cannot handle a pair of trivial questions.

Meanwhile, what's the 3rd question? There were only two at the bottom of the page. You mean your perpendicular scenario above? I answered that in post 23.

I pointed out several things in post 25. The beginning starts off with some kind of analogy of a car on a road with water on it. Light, to the surprise of the scientists of the late 19th century, does not behave like it travels through a medium, so not sure what all that was trying to prove. I pretty much skipped to the bottom where I found bad physics concerning fiberoptic gyroscopes and then a section at the bottom that asserted that SR cannot handle a pair of trivial questions.

Meanwhile, what's the 3rd question? There were only two at the bottom of the page. You mean your perpendicular scenario above? I answered that in post 23.
I see your answered that in post 23 . According to your formula: w = (u+v)/(1+uv/c^2) , the calculation is correct.

“The beginning starts off with some kind of analogy of a car on a road with water on it.”
This is the bedding for introducing light. If you can look down patiently, you will understand. But you don't have to look at it at all. What the article wants to explain is the influence of gravity on light, just like LIGO observed that weak gravitational waves can affect the propagation of light.

Forget the fiber optic gyroscope. Let's continue to focus on my questions.

Third scene:
Earth---------------------------------------A--->
.|....................................................0.9C
.|
.|
.|
.|
\/ B 0.9C

This is a simple hypotenuse.
Work out the naive speeds relative to each other, (using the hypotenuse formula) then apply the relativistic velocity addition formula.

Let's continue to focus on my questions.
Let's focus on you getting to where you can answer your own questions, rather than using us as a library.

Try the above calculation. Break it down into two components.

Fourth scene:
A------------------L=9LS---------------->Earth<------------------L=9LS-----------------------B
v=0.9C
............................................................................................................v=0.9C

With Earth as the reference, A's time goes slower, and with Earth as the reference, B's time goes slower. So when they meet, the time is same.
but
With A as the reference, B's time goes slower, and with B as the reference, A's time goes slower. When they meet, please tell me whose time is slower?

This is a simple hypotenuse.
Work out the naive speeds relative to each other, (using the hypotenuse formula) then apply the relativistic velocity addition formula.
Yes, it is very simple, this calculation process comes from classical physics.

Yes, it is very simple, this calculation process comes from classical physics.
No. From straight math.
Figure out the hypotenuse length of a right triangle with sides .9
That will get you the classical receding velocity.
Then apply the formula.

No. From straight math.
Figure out the hypotenuse length of a right triangle with sides .9
That will get you the classical receding velocity.
Then apply the formula.
OK, are you sure you're right?

First scene:
<--A----------------------------------Earth---------------------------------------B--->
-0.9C............................................................................................................0.9C

Second scene:
Earth---------------------------------------A--->
----------------------------------------------B--->
.............................................................0.9C
0.0c.
In the first case we are finding the velocity between A and B as measured by A or B, so the directions of the velocities must be taken as measured by which of the two we are making the measurement from.
In the case of A we get:
The Earth is moving to the right relative to A and B is moving to the right relative to the. If we assign motion to the right as positive then both u and v are positive.
In your second scene The Earth, as measured from A is moving to to the Left while B is still moving to the right relative to the Earth, for A, the scene looks like.
<--Earth---------------------------A
0.9c
-------------------------------------B-->
0.9c

u is -0.9c and v is 0.9c
Which yields (-0.9c)+0.9c)/(1+(-(0.9c)(0.9c)/c^2) = 0c

Third scene We will look at it from the rest frame for A where you get:
0.9c
<--Earth-----------------A
|
|
|
\/ B

Since the Earth-B pair share a 0.9c to the left velocity, this pair is time dilated by a factor of 0.436 as measured by A. This includes the vertical speed of B with respect to the Earth. That means that B vertical speed component is 0.392c (which is why I drew the Earth-B line shorter than the Earth-A line.
Now we can just do vector velocity addition to get sqrt ((0.9c)^2+ (0.392c)^2) = 0.9817c